Respuesta :
The pH of the solution prepared by mixing NaOH and [tex]\rm HNO_3[/tex] has been 1.69.
The moles of NaOH can be given by:
Moles = Molarity × Volume (L)
- Moles of NaOH = 0.1 M × 0.02 L
Moles of NaOH = 0.002 mol
- Moles of [tex]\rm HNO_3[/tex] = 0.1 M × 0.03 L
Moles of [tex]\rm HNO_3[/tex] = 0.003 mol
Since, [tex]\rm HNO_3[/tex] has been remaining in the solution after neutralization, the moles of [tex]\rm HNO_3[/tex] has been :
[tex]\rm HNO_3[/tex] = 0.003 - 0.002 mol
[tex]\rm HNO_3[/tex] = 0.001 mol.
The moles of [tex]\rm HNO_3[/tex] has been equivalent to the moles of hydrogen ion.
The hydrogen ion has been 0.001 mol.
The concentration of Hydrogen ions can be given by:
Hydrogen ion concentration = [tex]\rm \dfrac{Moles}{Volume(L)}[/tex]
Hydrogen ion concentration = [tex]\rm \dfrac{0.001}{0.03\;+\;0.02}[/tex]
Hydrogen ion concentration = 0.02 M.
pH can be defined as the negative log of the hydrogen ion concentration.
pH = - log ([tex]\rm H^+[/tex])
pH = - log (0.02)
pH = 1.69
The pH of the solution prepared by mixing NaOH and [tex]\rm HNO_3[/tex] has been 1.69.
For more information about the pH of the solution, refer to the link:
https://brainly.com/question/4975103
